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Question Number 178967 by Tawa11 last updated on 23/Oct/22

Find the greatest coefficient in expansion of: (3x − 2)^(25)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{coefficient}\:\mathrm{in}\:\mathrm{expansion}\:\mathrm{of}:\:\:\:\left(\mathrm{3x}\:\:\:−\:\:\:\mathrm{2}\right)^{\mathrm{25}} \\ $$

Answered by mr W last updated on 23/Oct/22

a_n =C_n ^(25) 3^n (−2)^(25−n) n=2k+1 for positive coef. a_(2k+1) =C_(2k+1) ^(25) 3^(2k+1) (−2)^(25−2k−1) =C_(2k+1) ^(25) 3^(2k+1) 2^(24−2k) a_(2(k+1)+1) =C_(2k+3) ^(25) 3^(2k+3) 2^(22−2k) a_(2(k+1)+1) <a_(2k+1) C_(2k+3) ^(25) 3^(2k+3) 2^(22−2k) <C_(2k+1) ^(25) 3^(2k+1) 2^(24−2k) 3^2 ×((25!)/((2k+3)!(25−2k−3)!))<2^2 ×((25!)/((2k+1)!(25−2k−1)!)) (9/((2k+3)(2k+2)))<(4/((24−2k)(23−2k))) 10k^2 −443k+2472<0 ((443−(√(97369)))/(20))<k<((443+(√(97369)))/(20)) ⇒7≤k≤37 ⇒maximum coef. is at k=6, i.e. n=13 a_(max) =a_(13) =C_(13) ^(25) 3^(13) 2^(12)

$${a}_{{n}} ={C}_{{n}} ^{\mathrm{25}} \mathrm{3}^{{n}} \left(−\mathrm{2}\right)^{\mathrm{25}−{n}} \\ $$$${n}=\mathrm{2}{k}+\mathrm{1}\:{for}\:{positive}\:{coef}. \\ $$$${a}_{\mathrm{2}{k}+\mathrm{1}} ={C}_{\mathrm{2}{k}+\mathrm{1}} ^{\mathrm{25}} \mathrm{3}^{\mathrm{2}{k}+\mathrm{1}} \left(−\mathrm{2}\right)^{\mathrm{25}−\mathrm{2}{k}−\mathrm{1}} ={C}_{\mathrm{2}{k}+\mathrm{1}} ^{\mathrm{25}} \mathrm{3}^{\mathrm{2}{k}+\mathrm{1}} \mathrm{2}^{\mathrm{24}−\mathrm{2}{k}} \\ $$$${a}_{\mathrm{2}\left({k}+\mathrm{1}\right)+\mathrm{1}} ={C}_{\mathrm{2}{k}+\mathrm{3}} ^{\mathrm{25}} \mathrm{3}^{\mathrm{2}{k}+\mathrm{3}} \mathrm{2}^{\mathrm{22}−\mathrm{2}{k}} \\ $$$${a}_{\mathrm{2}\left({k}+\mathrm{1}\right)+\mathrm{1}} <{a}_{\mathrm{2}{k}+\mathrm{1}} \\ $$$${C}_{\mathrm{2}{k}+\mathrm{3}} ^{\mathrm{25}} \mathrm{3}^{\mathrm{2}{k}+\mathrm{3}} \mathrm{2}^{\mathrm{22}−\mathrm{2}{k}} <{C}_{\mathrm{2}{k}+\mathrm{1}} ^{\mathrm{25}} \mathrm{3}^{\mathrm{2}{k}+\mathrm{1}} \mathrm{2}^{\mathrm{24}−\mathrm{2}{k}} \\ $$$$\mathrm{3}^{\mathrm{2}} ×\frac{\mathrm{25}!}{\left(\mathrm{2}{k}+\mathrm{3}\right)!\left(\mathrm{25}−\mathrm{2}{k}−\mathrm{3}\right)!}<\mathrm{2}^{\mathrm{2}} ×\frac{\mathrm{25}!}{\left(\mathrm{2}{k}+\mathrm{1}\right)!\left(\mathrm{25}−\mathrm{2}{k}−\mathrm{1}\right)!} \\ $$$$\frac{\mathrm{9}}{\left(\mathrm{2}{k}+\mathrm{3}\right)\left(\mathrm{2}{k}+\mathrm{2}\right)}<\frac{\mathrm{4}}{\left(\mathrm{24}−\mathrm{2}{k}\right)\left(\mathrm{23}−\mathrm{2}{k}\right)} \\ $$$$\mathrm{10}{k}^{\mathrm{2}} −\mathrm{443}{k}+\mathrm{2472}<\mathrm{0} \\ $$$$\frac{\mathrm{443}−\sqrt{\mathrm{97369}}}{\mathrm{20}}<{k}<\frac{\mathrm{443}+\sqrt{\mathrm{97369}}}{\mathrm{20}} \\ $$$$\Rightarrow\mathrm{7}\leqslant{k}\leqslant\mathrm{37} \\ $$$$\Rightarrow{maximum}\:{coef}.\:{is}\:{at}\:{k}=\mathrm{6},\:{i}.{e}.\:{n}=\mathrm{13} \\ $$$${a}_{{max}} ={a}_{\mathrm{13}} ={C}_{\mathrm{13}} ^{\mathrm{25}} \mathrm{3}^{\mathrm{13}} \mathrm{2}^{\mathrm{12}} \\ $$

Commented by Tawa11 last updated on 23/Oct/22

God bless you sir. I really appreciate your time sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir}. \\ $$

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